MathsCBSE Class 10 · Basic · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic sample paper 2024-25: question 34

From the official CBSE sample question paper for 2024-25, section D, question 34, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34 · 5 marks · Long answerAreas Related to Circles

The perimeter of sector OACB of the circle centred at O and of radius 24, is 73.12 cm.

(i) Find the central angle $∠ A O B$ .
(ii) Find the area of the minor segment ACB. (Use $π = 3.14$ and $3 = 1.73$ )

Figure in the paper: A circle with centre O and radius 24. Points A and B on the circle are joined to O, forming sector OACB with arc ACB (C on the minor arc). Chord AB is drawn. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

The perimeter of a sector is two radii plus the arc length.

Hint 2

With $∠ A O B = 6 0^{\circ}$ and $O A = O B$ , triangle OAB is equilateral. Segment area = sector area − triangle area.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Perimeter of the sector $= 2 r + \frac{θ}{36 0 ^{\circ}} \times 2 π r$ , so:
$2 (24) + \frac{2 \times 3.14 \times 24 \times θ}{360} = 73.12$
1 mark
$\frac{150.72 θ}{360} = 73.12 - 48 = 25.12 \Rightarrow θ = \frac{25.12 \times 360}{150.72} = 6 0^{\circ}$ .
So $∠ A O B = 6 0^{\circ}$ .
1 mark
(ii) Since $O A = O B$ and $∠ A O B = 6 0^{\circ}$ , $△ O A B$ is equilateral with side 24 cm.
Area of minor segment = area of sector − area of $△ O A B$
$= (\frac{60}{360} \times 3.14 \times 24 \times 24 - \frac{1.73}{4} \times 24 \times 24)$ cm²
2 marks
$= (301.44 - 249.12)$ cm² $= 52.32$ cm²
1 mark

Answer: (i) $6 0^{\circ}$ ; (ii) 52.32 cm²

Where the marks are

Writes the perimeter equation $2 r + \frac{2 π r θ}{360} = 73.12$ (1 mark)
Solves to get $θ = 6 0^{\circ}$ (1 mark)
Segment area = sector area − equilateral triangle area, with correct substitution (2 marks)
Area = 52.32 cm² (1 mark)

Revise the chapter

Chapter: Areas Related to Circles (notes, formulas, where marks are lost and every past question)
Topic: Area of a Sector of a Circle

Practise this chapterBoard-style questions from Areas Related to Circles, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2024-25 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic sample paper 2024-25: question 34

Hints

Step-by-step solution

Revise the chapter

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